Reward

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6336    Accepted Submission(s): 1958

Problem Description Dandelion’s uncle is a boss of a factory. As the spring festival is coming , he wants to distribute rewards to his workers. Now he has a trouble about how to distribute the rewards.

The workers will compare their rewards ,and some one may have demands of the distributing of rewards ,just like a’s reward should more than b’s.Dandelion’s unclue wants to fulfill all the demands, of course ,he wants to use the least money.Every work’s reward will be at least 888 , because it’s a lucky number.  

Input One line with two integers n and m ,stands for the number of works and the number of demands .(n<=10000,m<=20000)

then m lines ,each line contains two integers a and b ,stands for a’s reward should be more than b’s.  

Output For every case ,print the least money dandelion ‘s uncle needs to distribute .If it’s impossible to fulfill all the works’ demands ,print -1.  

Sample Input

2 1 1 2 2 2 1 2 2 1  

Sample Output

1777 -1  

Author dandelion  

Source
曾是惊鸿照影来  

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Note

错了n次。。因为想错了。

一开始用并查集判断成环来判断是否冲突，后来发现这样是不对了

比如 

3 3

1 2

2 3

1 3

按照3->2->1是正确的。

给大家我测试的数据：4 4 1 2 2 3 1 4 3 4     3558

3 2 1 2 1 3 2665

3 1 1 2 2665

我用的邻接vector 来表示边。

#include <stdio.h>
#include <string.h>
#include <vector>
#include <algorithm>
#include <queue>
using namespace std;
int in[10005],add[10005];//in表示入度,add表示奖励
vector<int>map[10005];
int top_sort(int n)
{
	queue<int>s;
	for(int i=1;i<=n;i++)
	if(in[i]==0)
	s.push(i);
	int count=0;
	while(!s.empty())
	{
		int x=s.front();
		s.pop();
		count++;
		for(int i=0;i<map[x].size();i++)
		{
			in[map[x][i]]--;
			if(in[map[x][i]]==0)
			{
				s.push(map[x][i]);
				add[map[x][i]]=max(add[x]+1,add[map[x][i]]);//这里需要好好领悟。
			}
		}
	}
	if(count!=n)
	return -1;
	int sum=0;
	for(int i=1;i<=n;i++)
	sum+=add[i]+888;
	return sum;
}
int main()
{
	int n,m;
	while(~scanf("%d %d",&n,&m))
	{
		memset(in,0,sizeof(in));
		memset(map,0,sizeof(map));
		memset(add,0,sizeof(add));
		for(int i=0;i<m;i++)
		{
			int a,b;
			scanf("%d %d",&a,&b);
			map[b].push_back(a);
			in[a]++;
		}
		printf("%d\n",top_sort(n));
	}
	return 0;
}